Question

If $1+6+11+16+...+x=148$, then the value of $x$ is

• A. $36$
• B. $35$
• C. $-36$
• D. None of these

Answer 1

The correct option is A $36$

Given series is
$1+6+11+16+.....+x=148$
Here, $a=1,d=6-1=11-6=16-11=5$ and $S_n=148$
$\because S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$\therefore 148=\frac{n}{2}[2\times 1+(n-1\left)5\right]$
$\Rightarrow 296=2n+5n^2-5n$
$\Rightarrow 296=5n^2-3n$
$\Rightarrow 5n^2-3n-296=0$
$\Rightarrow (n-8)(5n+37)=0$
$\Rightarrow n=8,n=\frac{-37}{5}$
$\therefore n=8$
$(\because n$ cannot be negative)
Now, $T_n=a+(n-1)d$
$x=1+(8-1)\times 5$
$\Rightarrow x=1+35\Rightarrow x=36$.